Chapter MS, Problem 4PQ

Interpretation Introduction

Interpretation:

The pair of molecules that are geometrically similar has to be determined.

Concept Introduction:

VSEPR or valence shell electron pair repulsion theory predicts the shape of the molecules with the use of arrangement of electrons around the central metal atom. Its postulates are as follows:

1. The shape of the molecules is governed by the repulsion between electron pairs situated in the outermost or valence shell of the atoms.

2. A distortion is observed in the shape of the molecules if lone pairs are present in it. These are the electron pairs that do not participate in the chemical bonding between the atoms. The electron pairs that take part in bond formation are known as bond pairs.

3. The order of repulsion of electron pairs is as follows:

  $\text{lp}-\text{lp repulsion}>\text{lp}-\text{bp repulsion }>\text{bp}-\text{bp repulsion}$

4. Multiple bonds create more repulsion as compared to single bonds.

5. The following table shows the various molecular shapes that match to different number of electron pairs:

$\begin{array}{ccc}\begin{array}{l}\text{Steric number or}\\ \text{number of}\\ \text{electron groups}\end{array}& \begin{array}{l}\text{Number of }\\ \text{lone pairs}\end{array}& \text{Molecular geometry}\\ 2& 0& \text{Linear}\\ 3& 0& \text{Trigonal planar}\\ 3& 1& \text{Bent}\\ 4& 0& \text{Tetrahedral}\\ 4& 1& \text{Trigonal pyramid}\\ 4& 2& \text{Bent}\\ 5& 0& \text{Trigonal bipyramidal}\\ 5& 1& \text{See}-\text{saw}\\ 5& 2& \text{T}-\text{structure}\\ 5& 3& \text{Linear}\\ 6& 0& \text{Octahedral}\\ 6& 1& \text{Square pyramidal}\\ 6& 2& \text{Square planar}\end{array}$

The geometry of the molecule is predicted with the help of steric number. The formula to calculate the steric number is as follows:

  $\text{Steric number}=\left[\begin{array}{c}\left(\text{Atoms bonded to the central atom}\right)+\\ \left(\text{Number of lone pairs on the central atom}\right)\end{array}\right]$        (1)

The least electronegative atom is usually taken to be the central atom.

Answer to Problem 4PQ

Option (D) is the correct one.

Explanation of Solution

Reason for correct option:

(D) Substitute 2 for atoms bonded to the central atom and 1 for number of lone pairs on the central atom in equation (1) to calculate the steric number in ${\text{SO}}_{\text{2}}$.

  $\begin{array}{c}{\text{Steric number in SO}}_2=\text{2}+\text{1}\\ =\text{3}\end{array}$

So the number of electron pairs in ${\text{SO}}_{\text{2}}$ is 3 and one lone pair is also present on it. Therefore its geometry is bent according to the VSEPR theory.

Substitute 2 for atoms bonded to the central atom and 1 for number of lone pairs on the central atom in equation (1) to calculate the steric number in ${\text{O}}_3$.

  $\begin{array}{c}{\text{Steric number in O}}_3=\text{2}+\text{1}\\ =\text{3}\end{array}$

So the number of electron pairs in ${\text{O}}_3$ is 3 and one lone pair is also present on it. Therefore its geometry is bent according to the VSEPR theory.

Therefore ${\text{SO}}_{\text{2}}$ and ${\text{O}}_3$ are geometrically similar. Hence option (D) is correct.

Reason for incorrect option:

(A) Substitute 2 for atoms bonded to the central atom and 1 for number of lone pairs on the central atom in equation (1) to calculate the steric number in ${\text{SO}}_{\text{2}}$.

  $\begin{array}{c}{\text{Steric number in SO}}_2=\text{2}+\text{1}\\ =\text{3}\end{array}$

So the number of electron pairs in ${\text{SO}}_{\text{2}}$ is 3 and one lone pair is also present on it. Therefore its geometry is bent according to the VSEPR theory.

Substitute 2 for atoms bonded to the central atom and 0 for number of lone pairs on the central atom in equation (1) to calculate the steric number in ${\text{CO}}_{\text{2}}$.

  $\begin{array}{c}{\text{Steric number in CO}}_2=\text{2}+\text{0}\\ =\text{2}\end{array}$

So the number of electron pairs in ${\text{CO}}_{\text{2}}$ is 2 and no lone pair is also present on it. Therefore its geometry is linear according to the VSEPR theory.

(B) Substitute 3 for atoms bonded to the central atom and 1 for number of lone pairs on the central atom in equation (1) to calculate the steric number in ${\text{PH}}_3$.

  $\begin{array}{c}{\text{Steric number in PH}}_3=\text{3}+\text{1}\\ =\text{4}\end{array}$

So the number of electron pairs in ${\text{PH}}_3$ is 4 and one lone pair is also present on it. Therefore its geometry is trigonal pyramidal according to the VSEPR theory.

Substitute 3 for atoms bonded to the central atom and 0 for number of lone pairs on the central atom in equation (1) to calculate the steric number in ${\text{BF}}_3$.

  $\begin{array}{c}{\text{Steric number in BF}}_3=\text{3}+\text{0}\\ =\text{3}\end{array}$

So the number of electron pairs in ${\text{BF}}_3$ is 3 and no lone pair is also present on it. Therefore its geometry is trigonal planar according to the VSEPR theory.

(C) Substitute 2 for atoms bonded to the central atom and 0 for number of lone pairs on the central atom in equation (1) to calculate the steric number in ${\text{CO}}_{\text{2}}$.

  $\begin{array}{c}{\text{Steric number in CO}}_2=\text{2}+\text{0}\\ =\text{2}\end{array}$

So the number of electron pairs in ${\text{CO}}_{\text{2}}$ is 2 and no lone pair is also present on it. Therefore its geometry is linear according to the VSEPR theory.

Substitute 2 for atoms bonded to the central atom and 2 for number of lone pairs on the central atom in equation (1) to calculate the steric number in ${\text{OF}}_{\text{2}}$.

  $\begin{array}{c}{\text{Steric number in OF}}_{\text{2}}=\text{2}+\text{2}\\ =\text{4}\end{array}$

So the number of electron pairs in ${\text{OF}}_{\text{2}}$ is 4 and two lone pairs are also present on it. Therefore its geometry is bent according to the VSEPR theory.

Hence option (A), (B) and (C) are incorrect.

Conclusion

The pair of molecules that are geometrically similar is ${\text{SO}}_{\text{2}}$ and ${\text{O}}_{\text{3}}$. Hence, the correct option is (D).